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madam [21]
3 years ago
12

Please help me, need an expert.

Mathematics
2 answers:
n200080 [17]3 years ago
5 0
The second one, 7 < x < 13
oksian1 [2.3K]3 years ago
5 0
Domain is the numbers you can use
since this is a linear function, the endpoints should be the min and max
range is the numbers y ou get from nputing the domain
x is domain  y is range
so
2 to 5
subsitut 2 for x
y=2(2)+3
y=4+3
y=7

subsitute 5
y=2(5)+3
y=10+3
y=13
so range is
7≤y≤13
2nd option
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Answer:Answer: A negative correlation shows the relationship between two variables.

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Kofi is 9 years old. Kwaku's age is two-thirds Kofi's age
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<h3>3 : 2 : 4</h3>

<h2>CALCULATIONS</h2>

<h3>Let;</h3>

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A garden is in the shape of a rectangle 85m by 50m
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7 0
4 years ago
Read 2 more answers
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
amm1812

Answer:

6546 students would need to be sampled.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.

This means that n = 200, \pi = \frac{118}{200} = 0.59

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled?

n students would need to be sampled, and n is found when M = 0.01. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.01 = 1.645\sqrt{\frac{0.59*0.41}{n}}

0.01\sqrt{n} = 1.645\sqrt{0.59*0.41}

\sqrt{n} = \frac{1.645\sqrt{0.59*0.41}}{0.01}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.59*0.41}}{0.01})^2

n = 6545.9

Rounding up:

6546 students would need to be sampled.

3 0
3 years ago
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